Ex 13.4, 8 - A random variable X has probability distribution - Probability distribution

Ex 13.4, 8 - Chapter 13 Class 12 Probability - Part 2
Ex 13.4, 8 - Chapter 13 Class 12 Probability - Part 3
Ex 13.4, 8 - Chapter 13 Class 12 Probability - Part 4
Ex 13.4, 8 - Chapter 13 Class 12 Probability - Part 5

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Question 8 A random variable X has the following probability distribution: Determine (i) k Since X is a random variable , its Sum of Probabilities is equal to 1 0﷮7﷮𝑃(𝑋)﷯ = 1 P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1 9k + 10k2 = 1 10k2 + 9k – 1 = 0 10k2 + 10k – k – 1 = 0 10k2 + 10k – k – 1 = 0 10k(k + 1) – 1(k + 1) = 0 (10k – 1)(k + 1) = 0 So, k = 1﷮10﷯ & k = –1 But k is probability, So, it cannot be negative, Hence, k = 𝟏﷮𝟏𝟎﷯ Question 8 A random variable X has the following probability distribution: Determine (ii) P(X < 3) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 2k = 3k = 3 × 1﷮10﷯ = 𝟑﷮𝟏𝟎﷯ Question 8 A random variable X has the following probability distribution: Determine (iii) P(X > 6) P(X > 6) = P(X = 7) = 7k2 + k = 7 1﷮10﷯﷯﷮2﷯+ 1﷮10﷯ = 7 × 1﷮100﷯+ 1﷮10﷯ = 7﷮100﷯+ 1﷮10﷯ = 𝟏𝟕﷮𝟏𝟎𝟎﷯ Question 8 A random variable X has the following probability distribution: Determine (iv) P(0 < X < 3) P(0 < X < 3) = P(X = 1) + P(X = 2) = k + 2k = 3k = 3 × 1﷮10﷯ = 𝟑﷮𝟏𝟎﷯

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo